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Svetlanka [38]
3 years ago
9

Which equation correctly relates mechanical energy, thermal energy, and total energy when there is friction present in a system?

​
Physics
2 answers:
AleksAgata [21]3 years ago
6 0

Answer:

Ethermal = Etotal − ME

a p e x

fenix001 [56]3 years ago
4 0

Answer:ME=E TOTAL - E THERMAL

Explanation:JUST DID IT ON APEX

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A car is driving away from a crosswalk. The formula d = t 2 + 2 t expresses the car's distance from the crosswalk in feet, d , i
Ede4ka [16]

Answer:

1) No, the car does not travel at constant speed.

2) V = 9 ft/s

3) No, the car does not travel at constant speed.

4) V = 5.9 ft/s

Explanation:

In order to know if the car is traveling at constant speed we need to derive the given formula. That way we get speed as a function of time:

V(t) = 2*t + 2   Since the speed depends on time, the speed is not constant at any time.

For the average speed we evaluate the formula for t=2 and t=5:

d(2) = 8 ft     and      d(5) = 35 ft

V_{2-5}=\frac{d(5)-d(2)}{5-2}=9 ft/s

Again, for the average speed we evaluate the formula for t=1.8 and t=2.1:

d(1.8) = 6.84 ft     and      d(2.1) = 8.61 ft

V_{1.8-2.1}=\frac{d(2.1)-d(1.8)}{2.1-1.8}=5.9 ft/s

4 0
4 years ago
CALCULAR LA VELOCIDAD FINAL DE UN OBJETO, SI TIENE UNA VELOCIDAD
USPshnik [31]
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7 0
3 years ago
Three rocks with masses of 1 kg, 5 kg, and 10 kg fall from the same height.
Naily [24]
Answer:
The 10 kg rock has more inertia than the other two rocks.


Explanation
4 0
3 years ago
Read 2 more answers
A Venturi tube may be used as a fluid flowmeter. Suppose the device is used at a service station to measure the flow rate of gas
leonid [27]

Answer

given,

flow rate = p = 660 kg/m³

outer radius = 2.8 cm

P₁ - P₂ = 1.20 k Pa

inlet radius = 1.40 cm

using continuity equation

 A₁ v₁ = A₂ v₂

 π r₁² v₁ = π r₁² v₂

 v_1= \dfrac{r_1^2}{r_2^2} v_2

 v_1= \dfrac{1.4^2}{2.8^2} v_2

 v_1= 0.25 v_2

Applying Bernoulli's equation

 \Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)

 \Delta P = \dfrac{1}{2}\rho (v_2^2-(0.25 v_2)^2)

 \Delta P = \dfrac{1}{2}\rho v_2^2 (1 - 0.0625)

 v_2=\sqrt{\dfrac{2\Delta P}{\rho(1 - 0.0625)}}

 v_2=\sqrt{\dfrac{2\times 1200}{660 \times(1 - 0.0625)}}

       v₂ = 1.97 m/s

b) fluid flow rate

Q = A₂ V₂

Q = π (0.014)²  x 1.97

Q = 1.21 x 10⁻³ m³/s

5 0
4 years ago
A 1 200-kg automobile moving at 25 m/s has the brakes applied with a deceleration of 8.0 m/s2. How far does the car travel befor
Alja [10]

Answer:

Δx = 39.1 m

Explanation:

  • Assuming that deceleration keeps constant during the braking process, we can use one of the kinematics equations, as follows:

        v_{f} ^{2} - v_{o} ^{2} = 2* a * \Delta x (1)

        where  vf is the final velocity (0 in our case), v₀ is the initial velocity

        (25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance

        traveled since the brakes are applied.

  • Solving (1) for Δx, we have:

        \Delta x = \frac{-v_{o} ^{2} }{2*a} = \frac{-(25m/s)^{2}}{2*(-8.0m/s2} = 39.1 m (2)        

7 0
3 years ago
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